Calculate Buffer Effect of Adding HCl to Ethanoic Acid Solution

[josephcollins]

Hi ppl, could anyone give me a hint as to how to calculate this:

Calculate the effect of adding 10cm3 of hydrochloric acid of concentration 1M to 1dm3 of a buffer. The buffer is a solution of ethanoic acid(Ka=1.8 * 10^-5 M) at a concentration of 0.1M and sodium ethanoate at 0.1M.

Thanks for any help
Joe

 

[poolwin2001]

use Henderson-Hasselbalch equation:
$$pH=pK_a+log\frac{[A^-]}{[HA]}$$
if in doubt refer
https://www.physicsforums.com/showthread.php?t=44645
and the rest.

regards
P.S:Many posts have come on this topic.Moderator,why don't we make such a topic Sticky for a month or two?

 

[ravenprp]

To calculate the buffer effect, we need to first understand what a buffer is and how it works. A buffer is a solution that resists changes in pH when small amounts of acid or base are added to it. It does this by containing a weak acid and its conjugate base, or a weak base and its conjugate acid. In this case, the buffer solution is made up of ethanoic acid, a weak acid, and sodium ethanoate, its conjugate base.

To calculate the buffer effect of adding HCl to this solution, we can use the Henderson-Hasselbalch equation, which relates the pH of a solution to the pKa of the weak acid and the ratio of its conjugate base to the weak acid. The equation is:

pH = pKa + log([conjugate base]/[weak acid])

In this case, the pKa of ethanoic acid is 1.8 * 10^-5 M, and the concentrations of both the acid and its conjugate base are 0.1M. Plugging these values into the equation, we get:

pH = 1.8 * 10^-5 + log(0.1/0.1) = 4.75

This is the initial pH of the buffer solution. Now, let's see how adding 10cm3 of 1M HCl to 1dm3 of the buffer solution will affect the pH.

First, we need to calculate the moles of HCl added. Since the concentration of HCl is 1M, 10cm3, or 0.01dm3, of it will contain 0.01 moles of HCl.

Next, we need to calculate the new concentrations of ethanoic acid and sodium ethanoate. The volume of the solution has increased to 1.01dm3, but the concentrations of the acid and base remain the same. Therefore, the new concentrations will be:

[ethanoic acid] = (0.1 * 1dm3)/1.01dm3 = 0.099M
[sodium ethanoate] = (0.1 * 1dm3)/1.01dm3 = 0.099M

Now, we can plug these new values into the Henderson-Hasselbalch equation to calculate the new pH:

pH = 1.8 * 10^-5 + log(0.